5 Common Mistakes in Limit Problems (And How to Avoid Them)
Learn to spot and fix algebraic errors, misapplied rules, and domain issues that cost JEE aspirants valuable marks.
Why These Mistakes Cost JEE Aspirants
Based on analysis of JEE answer scripts, these 5 limit mistakes account for over 70% of all calculus errors in the exam. Understanding and avoiding them will give you:
- 3-8 additional marks in every JEE paper
- Faster problem-solving with error-checking habits
- Confidence in tricky limit problems
- Better time management by avoiding rework
Cancelling Zero Factors Incorrectly
Students often cancel $(x-a)$ factors before checking if they approach zero.
❌ Common Wrong Approach:
Find $\lim_{x \to 2} \frac{x^2-4}{x-2}$
Wrong: $\frac{x^2-4}{x-2} = x+2$ (cancelling directly) → Answer: 4
✅ Correct Solution:
Step 1: Factor numerator: $\frac{(x-2)(x+2)}{x-2}$
Step 2: Since $x \to 2$ but $x \neq 2$, we can cancel: $x+2$
Step 3: Now substitute: $2+2=4$
Correct Answer: 4 (but for the right reasons!)
🛡️ Prevention Strategy:
- Always check if factor approaches zero before cancelling
- Remember: $x \to a$ means $x$ gets close to $a$ but never equals $a$
- Use factorization properly for $\frac{0}{0}$ forms
Misapplying L'Hôpital's Rule
Using L'Hôpital's rule when conditions aren't satisfied or applying it incorrectly.
❌ Common Wrong Approach:
Find $\lim_{x \to 0} \frac{\sin x}{x}$
Wrong: Apply L'Hôpital directly without checking $\frac{0}{0}$ form
✅ Correct Solution:
Step 1: Check form: $\frac{\sin 0}{0} = \frac{0}{0}$ ✓
Step 2: Apply L'Hôpital: $\lim_{x \to 0} \frac{\cos x}{1}$
Step 3: Evaluate: $\frac{\cos 0}{1} = 1$
Correct Answer: 1
🛡️ Prevention Strategy:
- Only use L'Hôpital for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms
- Check differentiability of numerator and denominator
- Remember: Some limits are better solved by standard limits
Ignoring Domain Restrictions
Forgetting that limits consider approach from both sides and domain restrictions matter.
❌ Common Wrong Approach:
Find $\lim_{x \to 0} \sqrt{x}$
Wrong: Direct substitution gives $\sqrt{0} = 0$
✅ Correct Solution:
Step 1: Check domain: $\sqrt{x}$ defined only for $x \geq 0$
Step 2: Since domain is $[0, \infty)$, only right-hand limit exists
Step 3: $\lim_{x \to 0^+} \sqrt{x} = 0$
Correct Answer: Limit exists and equals 0
🛡️ Prevention Strategy:
- Always check function domain before finding limits
- Consider left-hand and right-hand limits separately
- Watch for square roots, logarithms, and rational functions
🚀 Limit Problem Checklist
Before Solving:
- Check function domain and restrictions
- Identify the limit form ($\frac{0}{0}$, $\frac{\infty}{\infty}$, etc.)
- Determine if left/right limits need separate treatment
- Look for standard limit forms
While Solving:
- Factorize before cancelling zero factors
- Verify L'Hôpital's rule conditions before applying
- Use algebraic manipulation for indeterminate forms
- Check both approaches for piecewise functions
Mistakes 4-5 Available in Full Version
Includes "Algebraic Manipulation Errors" and "Trigonometric Limit Confusions" with detailed solutions
📝 Spot the Mistake Exercise
Identify what's wrong with these limit solutions:
1. Find $\lim_{x \to 1} \frac{x^2-1}{x-1}$
Given Solution: $\frac{x^2-1}{x-1} = x+1 = 1+1 = 2$
2. Find $\lim_{x \to \infty} \frac{x+\sin x}{x}$
Given Solution: Apply L'Hôpital: $\lim_{x \to \infty} \frac{1+\cos x}{1}$ which doesn't exist.
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